【答案】
分析:(1)已知了b、c的值,即可確定拋物線的解析式,通過配方或用公式法即可求出其頂點E的坐標;
(2)在拋物線向下平移的過程中,拋物線的形狀沒有發(fā)生變化,所以b值不變,變化的只是c的值;可用c表示出A、B、C的坐標,若S
△BCE=S
△ABC,那么兩個三角形中BC邊上的高就應(yīng)該相等;可過E作EF∥BC,交x軸于F,根據(jù)平行線分線段成比例定理知AB=BF,由此可求出BF的長;易證得Rt△EDF∽Rt△COB,根據(jù)相似三角形所得到的成比例線段即可求出c的值,也就確定了拋物線的解析式,即可得到C、B的坐標,進而可用待定系數(shù)法求出直線BC的解析式;
(3)可設(shè)平移后拋物線的解析式為y=-(x-h)
2+k,與(2)的方法類似,也是通過作平行線,求出BF、DF的長,進而根據(jù)相似三角形來求出h、k的關(guān)系式,進而可根據(jù)E點在直線y=-4x+3上求出h、k的值,進而可確定平移后的拋物線解析式.
解答:解:(1)當b=2,c=3時,拋物線的解析式為y=-x
2+2x+3,即y=-(x-1)
2+4;
∴拋物線頂點E的坐標為(1,4)(2分)
(2)將(1)中的拋物線向下平移,則頂點E在對稱軸x=1上,有b=2,
∴拋物線的解析式為y=-x
2+2x+c(c>0);
∴此時,拋物線與y軸的交點為C(0,c),頂點為E(1,1+c);
∵方程-x
2+2x+c=0的兩個根為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/0.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/1.png)
,
∴此時,拋物線與x軸的交點為A(1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/2.png)
,0),B(1+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/3.png)
,0);
如圖,過點E作EF∥CB與x軸交于點F,連接CF,則S
△BCE=S
△BCF∵S
△BCE=S
△ABC,
∴S
△BCF=S
△ABC∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/4.png)
設(shè)對稱軸x=1與x軸交于點D,
則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/5.png)
由EF∥CB,得∠EFD=∠CBO
∴Rt△EDF∽Rt△COB,有
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/6.png)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/7.png)
結(jié)合題意,解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/8.png)
∴點
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/9.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/10.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/images11.png)
設(shè)直線BC的解析式為y=mx+n,則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/11.png)
,解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/12.png)
;
∴直線BC的解析式為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/13.png)
;(6分)
(3)根據(jù)題意,設(shè)拋物線的頂點為E(h,k),h>0,k>0;
則拋物線的解析式為y=-(x-h)
2+k,
此時,拋物線與y軸的交點為C,(0,-h
2+k),
與x軸的交點為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/14.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/15.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/16.png)
、
過點E作EF∥CB與x軸交于點F,連接CF,
則S
△BCE=S
△BCF;
由S
△BCE=2S
△AOC,
∴S
△BCF=2S
△AOC,得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/17.png)
;
設(shè)該拋物線的對稱軸與x軸交于點D;
則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/18.png)
;
于是,由Rt△EDF∽Rt△COB,有
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/19.png)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/20.png)
,即2h
3+(2k-3h
2)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/21.png)
-3hk=0,
(2h-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/22.png)
)(h-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/23.png)
)=0,
∵
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/24.png)
>h>0,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/25.png)
①,h=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/26.png)
(舍去),
∵點E(h,k)在直線y=-4x+3上,有k=-4h+3②
∴由①②,結(jié)合題意,解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/27.png)
有k=1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/28.png)
∴拋物線的解析式為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103957730923127/SYS201312111039577309231014_DA/29.png)
.(10分)
點評:本題著重考查了二次函數(shù)與坐標軸交點及頂點坐標的求法、二次函數(shù)圖象的平移、圖象面積的求法、平行線分線段成比例定理、相似三角形的判定和性質(zhì)等知識的綜合應(yīng)用能力,能力要求很高,難度較大.